Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{-2a - 2}{a^2 - 6a - 7} \div \dfrac{a - 5}{-a^2 + 2a + 35} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-2a - 2}{a^2 - 6a - 7} \times \dfrac{-a^2 + 2a + 35}{a - 5} $ First factor out any common factors. $z = \dfrac{-2(a + 1)}{a^2 - 6a - 7} \times \dfrac{-(a^2 - 2a - 35)}{a - 5} $ Then factor the quadratic expressions. $z = \dfrac {-2(a + 1)} {(a - 7)(a + 1)} \times \dfrac {-(a - 7)(a + 5)} {a - 5} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(a + 1) \times -(a - 7)(a + 5) } { (a - 7)(a + 1) \times (a - 5)} $ $z = \dfrac {2(a - 7)(a + 5)(a + 1)} {(a - 7)(a + 1)(a - 5)} $ Notice that $(a - 7)$ and $(a + 1)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {2\cancel{(a - 7)}(a + 5)(a + 1)} {\cancel{(a - 7)}(a + 1)(a - 5)} $ We are dividing by $a - 7$ , so $a - 7 \neq 0$ Therefore, $a \neq 7$ $z = \dfrac {2\cancel{(a - 7)}(a + 5)\cancel{(a + 1)}} {\cancel{(a - 7)}\cancel{(a + 1)}(a - 5)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $z = \dfrac {2(a + 5)} {a - 5} $ $ z = \dfrac{2(a + 5)}{a - 5}; a \neq 7; a \neq -1 $